1. Initial Setup:
Given the functional equation:
$$f(x+t) - f(x) = P(x)$$
for all $x \in \mathbb{R}$, where $P(x)$ is a polynomial with real coefficients and $t$ is a fixed real number.

2. Base Case:
Define $f(x)$ on the interval $[0, t)$ arbitrarily. Let $f(x)$ be any function on this interval.

3. Inductive Step:
Use the given functional equation to extend $f(x)$ to the entire real line. For $x \in [0, t)$, define:
$$f(x + t) = f(x) + P(x)$$
This extends $f(x)$ to the interval $[t, 2t)$.

4. Further Extension:
Continue this process iteratively. For $x \in [kt, (k+1)t)$ where $k$ is an integer, define:
$$f(x + t) = f(x) + P(x)$$
This extends $f(x)$ to the interval $[(k+1)t, (k+2)t)$.

5. Negative Direction:
Similarly, for $x \in [0, t)$, define:
$$f(x - t) = f(x) - P(x - t)$$
This extends $f(x)$ to the interval $[-t, 0)$.

6. General Form:
By induction, for any integer $k$, we have:
$$f(x + kt) = f(x) + \sum_{i=0}^{k-1} P(x + it)$$
and for negative $k$:
$$f(x + kt) = f(x) + \sum_{i=k}^{-1} P(x + it)$$

7. Conclusion:
The function $f(x)$ can be constructed piecewise using the given functional equation. The general solution is:
$$f(x) = g(x) + \sum_{i=0}^{\left\lfloor \frac{x}{t} \right\rfloor - 1} P(x - it)$$
where $g(x)$ is any function defined on the interval $[0, t)$.

The final answer is $\boxed{ f(x) = g(x) + \sum_{i=0}^{\left\lfloor \frac{x}{t} \right\rfloor - 1} P(x - it) }$ where $g(x)$ is any function defined on the interval $[0, t)$.